Mixing
Suppose an entire function $f$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero.
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
calculus complex-analysis
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
add a comment |Â
up vote
4
down vote
favorite
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
calculus complex-analysis
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
calculus complex-analysis
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
calculus complex-analysis
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
edited Jan 30 '14 at 3:14
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
asked Jan 29 '14 at 11:38
Shiquan
3,52911034
3,52911034
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01
add a comment |Â
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r in mathbbR$, so prove that $f$ cannot have two real zeros by using this together with the first part.
answered Jan 29 '14 at 11:44
user21820
35.8k440137
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
add a comment |Â
up vote
8
down vote
The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $gamma$, not necessarily closed, the integral $int_gammafracf'(z)f(z)dz$ represents the increase in argument of $f$ along $gamma$, i.e. how much the image of the path $gamma$ winds around the origin. And this last number need not be an integer multiple of $2pi i$.
Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:
We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc $=R$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $fcircgamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $fcircgamma$ will be approximately $6$ times that of $gamma$: $6(fracpi2) = 3pi$. Then we can examine $fcircgamma$ for the last segment of our curve and find it contributes an additional $pi$ to the argument. So we have a total increase of about $4pi$ in the argument of $f$ along $gamma$, which means there are $2$ zeros in the first quadrant.
So now to apply this idea to your problem. Let $gamma$ be a circle of radius $R$ around $0$. The integral $frac12pi iint_gammafracf'(z)f(z)dz$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $gamma$? First consider the increase around the upper half of the circle. $gamma(t) in mathbbR$ exactly when $t$ equals $0$ or $pi$, so by our hypothesis, these are the only times $fcircgamma$ can cross the real axis. So from $t$ equals $0$ to $pi$, the argument can only increase by exactly $pi$ or $-pi$. Then the same argument applies to the bottom half of the circle $gamma$. So as $gamma$ goes counter-clockwise around the origin, $fcircgamma$ will have an increase in argument of $2pi$, $0$, or $-2pi$. But the last possibility is excluded because our function has no poles.
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r in mathbbR$, so prove that $f$ cannot have two real zeros by using this together with the first part.
answered Jan 29 '14 at 11:44
user21820
35.8k440137
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
add a comment |Â
up vote
3
down vote
accepted
Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r in mathbbR$, so prove that $f$ cannot have two real zeros by using this together with the first part.
answered Jan 29 '14 at 11:44
user21820
35.8k440137
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r in mathbbR$, so prove that $f$ cannot have two real zeros by using this together with the first part.
answered Jan 29 '14 at 11:44
user21820
35.8k440137
Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r in mathbbR$, so prove that $f$ cannot have two real zeros by using this together with the first part.
answered Jan 29 '14 at 11:44
user21820
35.8k440137
answered Jan 29 '14 at 11:44
user21820
35.8k440137
answered Jan 29 '14 at 11:44
user21820
35.8k440137
answered Jan 29 '14 at 11:44
user21820
35.8k440137
35.8k440137
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
add a comment |Â
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
still confused...
– Shiquan
Jan 30 '14 at 3:41
still confused...
– Shiquan
Jan 30 '14 at 3:41
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem.
– user21820
Jan 31 '14 at 9:27
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
If $f(a)=0$ we have an $epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do?
– User Not Found
Oct 31 '16 at 6:55
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r in mathbbR to 0$. Then $f( r exp(it) ) = a r^2 exp(2it) + o(r^2)$ for any $t in mathbbR$, and to get the contradiction all we need is that it crosses the real axis at $t approx frac12 À$.
– user21820
Oct 31 '16 at 11:01
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
@ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $mathbbR$ has a stationary point and we can add a real constant to $f$ to get a contradiction.
– user21820
Oct 31 '16 at 11:07
add a comment |Â
up vote
8
down vote
The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $gamma$, not necessarily closed, the integral $int_gammafracf'(z)f(z)dz$ represents the increase in argument of $f$ along $gamma$, i.e. how much the image of the path $gamma$ winds around the origin. And this last number need not be an integer multiple of $2pi i$.
Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:
We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc $=R$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $fcircgamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $fcircgamma$ will be approximately $6$ times that of $gamma$: $6(fracpi2) = 3pi$. Then we can examine $fcircgamma$ for the last segment of our curve and find it contributes an additional $pi$ to the argument. So we have a total increase of about $4pi$ in the argument of $f$ along $gamma$, which means there are $2$ zeros in the first quadrant.
So now to apply this idea to your problem. Let $gamma$ be a circle of radius $R$ around $0$. The integral $frac12pi iint_gammafracf'(z)f(z)dz$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $gamma$? First consider the increase around the upper half of the circle. $gamma(t) in mathbbR$ exactly when $t$ equals $0$ or $pi$, so by our hypothesis, these are the only times $fcircgamma$ can cross the real axis. So from $t$ equals $0$ to $pi$, the argument can only increase by exactly $pi$ or $-pi$. Then the same argument applies to the bottom half of the circle $gamma$. So as $gamma$ goes counter-clockwise around the origin, $fcircgamma$ will have an increase in argument of $2pi$, $0$, or $-2pi$. But the last possibility is excluded because our function has no poles.
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
add a comment |Â
up vote
8
down vote
The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $gamma$, not necessarily closed, the integral $int_gammafracf'(z)f(z)dz$ represents the increase in argument of $f$ along $gamma$, i.e. how much the image of the path $gamma$ winds around the origin. And this last number need not be an integer multiple of $2pi i$.
Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:
We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc $=R$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $fcircgamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $fcircgamma$ will be approximately $6$ times that of $gamma$: $6(fracpi2) = 3pi$. Then we can examine $fcircgamma$ for the last segment of our curve and find it contributes an additional $pi$ to the argument. So we have a total increase of about $4pi$ in the argument of $f$ along $gamma$, which means there are $2$ zeros in the first quadrant.
So now to apply this idea to your problem. Let $gamma$ be a circle of radius $R$ around $0$. The integral $frac12pi iint_gammafracf'(z)f(z)dz$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $gamma$? First consider the increase around the upper half of the circle. $gamma(t) in mathbbR$ exactly when $t$ equals $0$ or $pi$, so by our hypothesis, these are the only times $fcircgamma$ can cross the real axis. So from $t$ equals $0$ to $pi$, the argument can only increase by exactly $pi$ or $-pi$. Then the same argument applies to the bottom half of the circle $gamma$. So as $gamma$ goes counter-clockwise around the origin, $fcircgamma$ will have an increase in argument of $2pi$, $0$, or $-2pi$. But the last possibility is excluded because our function has no poles.
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
add a comment |Â
up vote
8
down vote
up vote
8
down vote
The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $gamma$, not necessarily closed, the integral $int_gammafracf'(z)f(z)dz$ represents the increase in argument of $f$ along $gamma$, i.e. how much the image of the path $gamma$ winds around the origin. And this last number need not be an integer multiple of $2pi i$.
Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:
We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc $=R$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $fcircgamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $fcircgamma$ will be approximately $6$ times that of $gamma$: $6(fracpi2) = 3pi$. Then we can examine $fcircgamma$ for the last segment of our curve and find it contributes an additional $pi$ to the argument. So we have a total increase of about $4pi$ in the argument of $f$ along $gamma$, which means there are $2$ zeros in the first quadrant.
So now to apply this idea to your problem. Let $gamma$ be a circle of radius $R$ around $0$. The integral $frac12pi iint_gammafracf'(z)f(z)dz$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $gamma$? First consider the increase around the upper half of the circle. $gamma(t) in mathbbR$ exactly when $t$ equals $0$ or $pi$, so by our hypothesis, these are the only times $fcircgamma$ can cross the real axis. So from $t$ equals $0$ to $pi$, the argument can only increase by exactly $pi$ or $-pi$. Then the same argument applies to the bottom half of the circle $gamma$. So as $gamma$ goes counter-clockwise around the origin, $fcircgamma$ will have an increase in argument of $2pi$, $0$, or $-2pi$. But the last possibility is excluded because our function has no poles.
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $gamma$, not necessarily closed, the integral $int_gammafracf'(z)f(z)dz$ represents the increase in argument of $f$ along $gamma$, i.e. how much the image of the path $gamma$ winds around the origin. And this last number need not be an integer multiple of $2pi i$.
Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:
We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc $=R$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $fcircgamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $fcircgamma$ will be approximately $6$ times that of $gamma$: $6(fracpi2) = 3pi$. Then we can examine $fcircgamma$ for the last segment of our curve and find it contributes an additional $pi$ to the argument. So we have a total increase of about $4pi$ in the argument of $f$ along $gamma$, which means there are $2$ zeros in the first quadrant.
So now to apply this idea to your problem. Let $gamma$ be a circle of radius $R$ around $0$. The integral $frac12pi iint_gammafracf'(z)f(z)dz$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $gamma$? First consider the increase around the upper half of the circle. $gamma(t) in mathbbR$ exactly when $t$ equals $0$ or $pi$, so by our hypothesis, these are the only times $fcircgamma$ can cross the real axis. So from $t$ equals $0$ to $pi$, the argument can only increase by exactly $pi$ or $-pi$. Then the same argument applies to the bottom half of the circle $gamma$. So as $gamma$ goes counter-clockwise around the origin, $fcircgamma$ will have an increase in argument of $2pi$, $0$, or $-2pi$. But the last possibility is excluded because our function has no poles.
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
answered Jul 27 '14 at 19:40
PrimeRibeyeDeal
692514
692514
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
add a comment |Â
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
In that polynomial calculations, you meant $6(fracpi4)$ right?
– User Not Found
Oct 31 '16 at 20:49
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
Also how did you so easily say for last segment $pi$ arguement increases?
– User Not Found
Oct 31 '16 at 21:02
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f655831%2fsuppose-an-entire-function-f-is-real-if-and-only-if-z-is-real-prove-that-f%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
See math.stackexchange.com/q/1044732/96310 . In fact, we can prove that $f$ is a linear polynomial with real coefficients. See uwyo.edu/moorhouse/pub/farhad.pdf for the elementary solution.
– user1537366
Oct 31 '16 at 11:01